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The photon number in thermal blackbody radiation is not fixed, so $\mu =0$. With $x=E/(k_{B}T)$,The integral is a temperature-independent finite constant (its explicit value follows from the gamma-zeta integral). Differentiating at fixed volume gives

The Bose-Einstein distribution gives $n_i=[\exp ((E_i-\mu )/(k_{B}T))-1{]}^{-1}$, with chemical potential below the lowest energy when its discrete occupation is finite. In the continuum approximation,If the zero-energy ground state is macroscopically occupied, its discrete occupation $N_0$ must be added to the first integral; it contributes no energy to the second. This distinction is essential when applying the formulas to Bose-Einstein condensation.

Apply the Gaussian quadrature rule to the constant polynomial $1$. Exactness gives ${\sum }_{j=1}^n{A}_j=2$.

Use the Lagrange interpolation polynomialIts square has degree $2n-2$, so the exactness already proved givesThe positive weights of Gaussian quadrature follow because ${\ell }_j$ is a nonzero continuous polynomial. Thus these Gaussian quadrature weights are in fact positive, rather than merely nonnegative.

A degree-$n$ Legendre polynomial $P_n$ is orthogonal, for integration on $[-1,1]$, to every polynomial of degree at most $n-1$, and has $n$ distinct roots in $(-1,1)$. Divide $p$ by $P_n$:The Orthogonality of Legendre polynomials gives $\int p=\int r$. At every quadrature node $x_j$, $p(x_j)=r(x_j)$. The assumed exactness for $r$ therefore proves exactness up to degree $2n-1$, the characteristic property of Gaussian quadrature.

A homogeneous linear differential operator satisfies $L(e^{\epsilon }u)=e^{\epsilon }L(u)$, so the scaling $x\mapsto x$, $u\mapsto e^{\epsilon }u$ preserves every homogeneous linear equation. Its infinitesimal generator is $V=u{\partial }_u$.

On a patch with $u\ne 0$, take $s=\log |u|$, $t=x$. Then $V(s)=1$, $V(t)=0$. Set $w=ds/dt=u^{\prime }/u$. Since $u^{\prime \prime }/u=w^{\prime }+w^2$, division by $u$ reduces the second-order equation to the Riccati equationRecover a nonzero solution by $u=C\exp (\int wdt)$, allowing the sign in $C$. The zero solution is separate, and intervals containing zeros of other solutions require separate coordinate patches. The order reduction occurs because the transformed equation depends on $s^{\prime }$ but not on $s$ itself.

A Lie point symmetry is a one-parameter transformation of $(x,u)$ which maps the graph of a function solving the differential equation to another such graph of a function, wherever the transformed independent variable is invertible along the graph of a function. Its induced transformation on derivatives is the prolongation. Infinitesimally the criterion is ${\mathrm{pr}}^{(n)}V(\Delta )=0$ on $\Delta =0$. The derivative coefficients can be computed recursively from ${\eta }^{(0)}=\eta$ by ${\eta }^{(j)}=D_x{\eta }^{(j-1)}-u^{(j)}{D}_x\xi$. This expresses the requirement that the transformed derivatives still satisfy the equation.

The flow of the vector field is obtained by solvingDefine $g^{\epsilon }(x,u)$ to be this solution at parameter $\epsilon$. Uniqueness implies $g^0=\mathrm{id}$ and $g^{\epsilon +\delta }=g^{\epsilon }\circ g^{\delta }$ wherever both sides are defined: shifting the solution parameter gives the same initial-value problem. In particular $g^{-\epsilon }$ is the local inverse. This constructs the one-parameter group generated by the vector field, generally as a local group if the flow is not complete.

Use the nearly conserved oscillator energy $E=({\dot{x}}^2+x^2)/2$. The differential equation givesFor a leading-order orbit $x=A\cos \theta$, $\dot{x}=-A\sin \theta$, the cycle averages are $⟨{\dot{x}}^4⟩=3A^4/8$, $⟨x^3\dot{x}⟩=0$, $⟨x^4{\dot{x}}^2⟩=A^6/16$, and $⟨{\dot{x}}^2⟩=A^2/2$. The energy balance method therefore givesThe two nonzero simple roots give the leading limit cycles $x\simeq \cos (t+{\theta }_0)$ and $x\simeq \sqrt{2}\cos (t+{\theta }_0)$. Below amplitude $1$ the energy increases, between $1$ and $\sqrt{2}$ it decreases, and above $\sqrt{2}$ it increases. Hence the amplitude-$1$ cycle is stable and the amplitude-$\sqrt{2}$ cycle unstable for positive small $\epsilon$.

The linearized averaged radial exponents are $-\epsilon /2$ and $\epsilon$, respectively, consistent with nontrivial Floquet multipliers $e^{-\pi \epsilon +O({\epsilon }^2)}$ and $e^{2\pi \epsilon +O({\epsilon }^2)}$. The nonzero derivatives at these averaged zeros also justify persistence of nearby cycles for sufficiently small $\epsilon$.

The linearized perturbation obeys $\dot{\eta }=D\mathbf{f}(\mathbf{X}(t))\eta$, so its fundamental matrix satisfiesThe Floquet multipliers are the eigenvalues of the monodromy matrix $\Phi (T)$. Differentiating the orbit equation shows that $\dot{\mathbf{X}}(t)=\mathbf{f}(\mathbf{X}(t))$ is a solution of the variational equation. For a nonconstant periodic orbit this tangent solution is nonzero and periodic, giving one multiplier equal to $1$.

The Liouville formula for a fundamental matrix gives $(\det \Phi {)}^{\prime }=(\mathrm{tr}D\mathbf{f})\det \Phi$. Since $\det \Phi (0)=1$, the product of the two multipliers is $\exp ({\int }_0^T\nabla \cdot \mathbf{f}(\mathbf{X}(t))dt)$. The other multiplier is therefore exactly this exponential.

Expand the payoff as $(S_T-2K+K^2/S_T)1_{\{S_T>K\}}$. The same risk-neutral measure applies. For any real $p$, completing the square gives the truncated lognormal momentUsing $p=1,0,-1$ and discounting by $e^{-rT}$ givesThe inverse-price moment explains both the extra volatility factor and the shifted normal argument in the last term.

In the Black-Scholes model, absence of arbitrage and completeness give the price as the discounted expectation under the risk-neutral measure. Under that measure,The physical drift $\mu$ therefore does not enter the option price. PutFor the standard normal distribution function $\Phi$, $\mathbb{P}(S_T>K)=\Phi (d_2)$. Completing the square in the Gaussian density also gives $\mathbb{E}[S_T{1}_{\{S_T>K\}}]=S_0{e}^{rT}\Phi (d_1)$. Expanding the call payoff into these two truncated moments yields

A standard Brownian motion has $B_0=0$, continuous paths almost surely, and independent increments with $B_t-B_s\sim N(0,t-s)$ for $0\le s<t$. With a specified filtration, the increment is also independent of the information available at time $s$.

Put $s=\sqrt{n}/2$. The posterior is $\mathrm{Beta}(X+s,n-X+s)$, andThe denominator is positive for $n\ge 1$. For $n>4$, both parameters exceed one and the mode simplifies to $(X+s-1)/(n+\sqrt{n}-2)$. For $n=4$ this is the uniform-prior mode, exactly $X/n$; for $n=1,2$ the endpoint or symmetric cases also make the mode rule coincide with $X/n$. Otherwise it is generally a different rule. The posterior mean agrees with $X/n$ only at $X=n/2$.

To find the minimax estimator under quadratic loss, consider the posterior mean ${\delta }_s=(X+s)/(n+2s)$. Its risk function isFor $s^2=n/4$, the numerator is identically $n/4$, soThis constant-risk binomial proportion estimator is the unique Bayes estimator for its proper beta prior under quadratic loss. A competing rule's maximum risk is at least its risk averaged over that prior, which is at least this Bayes risk. The constant-risk rule attains the bound, so the posterior mean in (iii) is the unique minimax rule. Consequently any coincident rule is minimax too: the mean in (i) is the same rule when $n=4$, and the mean in (ii) is the same rule when $n=1$. No posterior mode listed here equals that mean rule for every observation, since at $X=0$ its value is positive whereas their endpoint modes differ from it. This accounts for the small-$n$ coincidences as well as the general answer.

The Jeffreys prior is proportional to $[\theta (1-\theta ){]}^{-1/2}$, hence is $\mathrm{Beta}(1/2,1/2)$. Its posterior is $\mathrm{Beta}(X+1/2,n-X+1/2)$, givingThe interior case is used only when $n\ge 2$. Generally neither rule equals $X/n$: the mean agrees only at $X=n/2$; the mode agrees at the endpoints and, in the interior, at $X=n/2$. For the small experiments $n=1,2$, all possible observations happen to be among these cases, so the whole mode rule then coincides with the maximum-likelihood estimator.

The uniform prior has $s=1$, so the posterior is $\mathrm{Beta}(X+1,n-X+1)$. Its mean and mode areThus the uniform-prior posterior mode is the maximum-likelihood estimator, including $X=0,n$. The mean shrinks toward $1/2$ and equals the maximum-likelihood estimator only when $X=n/2$.

The random clock is understood to be independent of the discrete-time Markov chain; without this hypothesis its law need not be determined by $K$ alone. Conditional on $N_t=n$, the transition matrix is $K^n$, soHence $q_{xy}=\lambda K_{xy}$ for $x\ne y$ and $q_{xx}=-\lambda (1-K_{xx})$. If $\pi K=\pi$, then $\pi P_t=\pi$ by the series, proving that the invariant distribution is unchanged.

Let ${\xi }_i$ be the independent exponential interarrival times of the Poisson process. The absorption time is $T_a={\sum }_{i=1}^{{\tau }_a}{\xi }_i$. By independence of the chain and clock and Tonelli theorem,This includes infinite expectations and absorption at time zero. The same argument proves the hinted random-sum identity for an independent nonnegative integer-valued count $M$.

A rate-$\lambda$ Poisson process is a counting process starting at zero, with right-continuous nondecreasing integer-valued paths, independent stationary increments, and $N_t-N_s\sim \mathrm{Pois}(\lambda (t-s))$ for $0\le s<t$. Its jumps have size one almost surely.

For independent Poisson processes $N,M$, increments of $N+M$ on disjoint intervals are independent, because all the corresponding increments of the two processes are jointly independent. On an interval of length $t$, multiplying their probability generating functions givesThus the increment is Poisson with mean $(\lambda +\mu )t$. The path requirements hold, and independent processes almost surely have no simultaneous jumps. Therefore $N+M$ is a Poisson process of rate $\lambda +\mu$.